Question: Solve for $x$ : $3x^2 + 24x + 36 = 0$
Dividing both sides by $3$ gives: $ x^2 + {8}x + {12} = 0 $ The coefficient on the $x$ term is $8$ and the constant term is $12$ , so we need to find two numbers that add up to $8$ and multiply to $12$ The two numbers $6$ and $2$ satisfy both conditions: $ {6} + {2} = {8} $ $ {6} \times {2} = {12} $ $(x + {6}) (x + {2}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x + 6) (x + 2) = 0$ $x + 6 = 0$ or $x + 2 = 0$ Thus, $x = -6$ and $x = -2$ are the solutions.